Question

How do you evaluate `arccos ( sin 50)`?

Answer 1

`(360n+-40)^o, n=0,+-1,+-2,+-3,...`

Explanation:

Had it been arc cos (cos a), the answer is that a.

Here, `sin 50^o=cos(360n+-40)^o, n=0,+-1,+-2,+-3,...`.

So, `arc cos (sin 50^o)=arc cos (cos(360n+-40)^o), n=0,+-1,+-2,+-3,...`

=`(360n+-40)^o, n=0,+-1,+-2,+-3,...`